Phase Diagrams, Response Functions and Fluctuations

Abstract

During the last decades emerging scientific attention has been directed towards the subject of phase transitions, namely the dramatic changeover in the properties of a macroscopic system upon variation of a thermodynamic variable (in most cases the temperature). In this chapter and in the subsequent three chapters, we shall deal with the description of some phases of solid state matter and describe the basic aspects of the microscopic mechanisms that control the transition between different phases. The key concept and the role of the order parameter will be introduced, first in the framework of the thermodynamic description and then in the framework of specific theories dealing with the microscopic dynamics that drive the crossover from one phase to the other.

Appendix 15.1 From Single Particle to Collective Response

The main lines of a general procedure which allows one to derive the response of a collective system (in the mean-field equilibrium state) to a time-dependent external perturbation are sketched in the following. The condition of linear response to both the external force and to the internal fluctuations shall be assumed. Furthermore interaction of bi-linear character will be considered while the local potential is somewhat arbitrary. The relevant advantage of this model is to include the fluctuations in the framework of the mean field approximation, where they are usually neglected. Thus the simplicity of the mean field approach is retained in spite of the inclusion of the fluctuations.¹⁴

The system is perturbed by a time-dependent perturbation $\varDelta F e^{-i \omega t}$, which modifies the local critical variable  $v_l$ according to

$$ v{^{\prime }}_{l}(t)=\varDelta v_{l}e^{-i \omega t}+v_{l}(t) $$

Weak perturbations and linear responses $\varDelta v_{l}=\chi ^0(\omega )\varDelta F_{l}$ are assumed, while $v_l(t)$ represents the spontaneous “local dynamics”. Finally the interactions among the local variables are described through a bilinear Hamiltonian

$$ \mathcal {H}= \frac{1}{2} \sum _{ll{^{\prime }}}\mathbf {I}_{ll{^{\prime }}}v_{l}v_{l{^{\prime }}}, $$

as for the Heisenberg exchange or for the harmonic lattice vibrations, for example. Then $v_{l}$ will not experience just the external perturbation $\varDelta F$ but also the feedback $\sum _{l{^{\prime }}}\mathbf {I}_{ll{^{\prime }}}v_{l{^{\prime }}}(t)$. Hence, $v_{l}$ will be globally perturbed by a term

$$ \varDelta F_{l}(t)=\varDelta F e^{-i \omega t}+ \sum _{l{^{\prime }}}\mathbf {I}_{ll{^{\prime }}}\varDelta v_{l{^{\prime }}}e^{-i \omega t}+ \sum _{l{^{\prime }}}\mathbf {I}_{ll{^{\prime }}}v_{l{^{\prime }}}(t) $$

where the last term describes the interactions involving the spontaneous fluctuations, which are present also in the absence of external stimulus. Within the mean field approximation one writes

$$\begin{aligned} \varDelta v_{l}=\chi ^0(\omega )\left[ \sum _{l{^{\prime }}}\mathbf {I}_{ll{^{\prime }}}\varDelta v_{l{^{\prime }}}+ \varDelta F_{l}\right] , \end{aligned}$$

where $\chi ^0(\omega )$ is the single-particle “bare” susceptibility. By resorting to the normal coordinates $\varDelta v_{\mathbf {q}}$ and expanding the stimulus in its Fourier components

$$\begin{aligned} \varDelta v_{l}=\frac{1}{\sqrt{N}}\sum _{\mathbf {q}}\varDelta v_{\mathbf {q}}e^{i\mathbf {q}\cdot \mathbf {R_{l}}} \end{aligned}$$

$$\begin{aligned} \varDelta F_{l}=\frac{1}{\sqrt{N}}\sum _{\mathbf {q}}\varDelta F_{\mathbf {q}}e^{i\mathbf {q}\cdot \mathbf {R_{l}}} \end{aligned}$$

from the previous equation one has

$$\begin{aligned} \sum _{\mathbf {q}}\varDelta v_{\mathbf {q}}e^{i\mathbf {q}\cdot \mathbf {R_{l}}} = \chi ^0(\omega ) \left[ \sum _{l{^{\prime }}}\mathbf {I}_{ll{^{\prime }}}\sum _{\mathbf {q}}\varDelta v_{\mathbf {q}}e^{i\mathbf {q}\cdot \mathbf {R_{l}{^{\prime }}}}+ \sum _{\mathbf {q}}\varDelta F_{\mathbf {q}}e^{i\mathbf {q}\cdot \mathbf {R_{l}}}\right] , \end{aligned}$$

which is rewritten in the form

$$\begin{aligned} \sum _{\mathbf {q}}\varDelta v_{\mathbf {q}}=\chi ^0(\omega ) \left[ \sum _{\mathbf {q}}\sum _{l{^{\prime }}}\mathbf {I}_{ll{^{\prime }}}\varDelta v_{\mathbf {q}}e^{i\mathbf {q}\cdot (\mathbf {R_{l{^{\prime }}}}-\mathbf {R_{l}})}+\sum _{\mathbf {q}}\varDelta F_{\mathbf {q}}\right] . \end{aligned}$$

Thus for each $\mathbf {q}$ one has

$$\begin{aligned} \varDelta v_{\mathbf {q}}= & {} \chi ^0(\omega ) \left[ \sum _{l{^{\prime }}}\mathbf {I}_{ll{^{\prime }}}\varDelta v_{\mathbf {q}}e^{i\mathbf {q}\cdot (\mathbf {R_{l{^{\prime }}}}-\mathbf {R_{l}})}+\varDelta F_{\mathbf {q}}\right] \\= & {} \chi ^0(\omega ) \left[ \mathbf {I_{q}}\varDelta v_{\mathbf {q}}+\varDelta F_{\mathbf {q}}\right] , \end{aligned}$$

where $\mathbf {I_q}=\sum _{l{^{\prime }}}\mathbf {I}_{ll{^{\prime }}}e^{i\mathbf {q}\cdot (\mathbf {R_{l{^{\prime }}}}-\mathbf {R_{l}})}$ is the Fourier transform of the spatially varying interaction. The collective susceptibility is

$$ \chi (\mathbf {q},\omega )=\frac{\varDelta v_{\mathbf {q}}}{\varDelta F_{\mathbf {q}}}, $$

and then

$$\begin{aligned} \chi (\mathbf {q}, \omega )=\frac{\chi ^0(\omega )}{1-\chi ^0(\omega )\mathbf {I_{q}}}. \end{aligned}$$

(A.15.1.1)

One observes that at $T_c$, for a given wavevector $\mathbf {q_c}$, so that $\chi ^0(0)I_{\mathbf {q_c}}\rightarrow 1$, a divergence of the static susceptibility $\chi (\mathbf {q}_c,0)$ occurs, implying the onset of a long range order with a modulation of the local variable determined by $\mathbf {q_c}$.

Let us refer to a system that can be described by a local quasi-harmonic potential of the form (b) sketched in Fig. 15.5.

Fig. 15.5

Sketchy forms of local strongly anharmonic potential (a) and of quasi-harmonic potential (b). a can describe order-disorder, relaxational-type systems while b quasi-harmonic vibrational lattices, displacive type systems (see Chap. 16)

In the case of no interaction the susceptibility would be the one of damped oscillator

$$ \chi ^0(\omega )=\chi ^0(0)\frac{\omega _{0}^2}{\omega _{0}^2-\omega ^2-i\gamma \omega }, $$

with $\omega _0$ fundamental frequency and $\gamma$ the damping parameter. The frequency $\omega _0$ can be considered slightly temperature dependent, for instance in the form $\omega _0\propto (a+ bT)$, $a \gg bT$ (quasi-harmonic approximation, see Problem 15.3). As a consequence of the interactions, from Eq. (A.15.1.1), one writes

$$ \chi (\mathbf {q},\omega )=\frac{1}{{\frac{1}{\chi ^{0}(\omega )}-\mathbf {I_{q}}}} =\frac{\omega _{0}^2 \chi ^{0}(0)}{\omega _{0}^2-\omega ^2-i\gamma \omega -\mathbf {I_{q}}\chi ^{0}(0)\omega _{0}^2}, $$

which preserves the same form of the susceptibility, with a frequency of the mode at wavevector $\mathbf {q}$ renormalized as

$$\begin{aligned} \omega _{\mathbf {q}}^2=\omega _{0}^2(1-\mathbf {I_{q}}\chi ^0(0)). \end{aligned}$$

(A.15.1.2)

In fact

$$ \chi (\mathbf {q},\omega )=\frac{\omega _{0}^2 \chi ^{0}(0)}{\omega _{\mathbf {q}}^2-\omega ^2-i\gamma \omega }= \frac{\omega _{\mathbf {q}}^2 \chi (\mathbf {q},0)}{\omega _{\mathbf {q}}^2-\omega ^2-i\gamma \omega } $$

and

$$ \frac{\chi (\mathbf {q},0)}{\chi ^0(0)}= \frac{\omega _{0}^2}{\omega _{\mathbf {q}}^2}. $$

When for $T\rightarrow T_c\,\,\,$ $\mathbf {I_{q}}\chi ^{0}(0)\rightarrow 1\,\,\,\,$,      $\omega _\mathbf {q}\rightarrow 0$.

For particles in a double-well strongly anharmonic potential (see (a) in the sketch above) (which with some modifications allows one to describe the parallel susceptibility of magnetic systems, see Chap. 17) the single particle susceptibility is given by Eq. (15.50) (relaxational behaviour) and the relaxation time $\tau$ is usually temperature dependent in the form $\tau \simeq \tau _0 exp(\varDelta E/k_BT)$. Then, from

$$ \chi (\mathbf {q},\omega ,T)=\frac{\chi ^{0}(0,T)}{1- \mathbf {I_{q}}\chi ^{0}(0,T) -i\omega \tau }, $$

a frequency $\omega _\mathbf {q}= i/\tau _\mathbf {q}$ goes to zero when according to Eq. (A.15.1.1) $\chi (\mathbf {q},0,T)$ goes to infinity and

$$ \tau _\mathbf {q}=\frac{\tau }{1- \mathbf {I_{q}}\chi ^{0}(0,T)} $$

also diverges (here the $T-$dependence is explicitly indicated).

Summarizing, it has been derived that for a given wave vector $\mathbf {q}$, for which as a function of the temperature $\mathbf {I_{q}}\chi ^{0}(0,T)$ approaches unity, a critical temperature $T_c(\mathbf {q})$ exists so that $\chi (\mathbf {q}, 0,T)\rightarrow \infty$ and $\omega _\mathbf {q}$ moves towards zero (see sketch below). Correspondingly the phase at $<v_l>=0$ becomes unstable against the generalized mode for which $T_c(\mathbf {q})$ takes the largest value.

The temperature dependence of the soft mode corresponds to the slowing down discussed at Sect. 15.4 in a different scenario. The detailed temperature dependence of the soft mode would need the knowledge of the microscopic Hamiltonian.

Since for fluctuations decoupled from the response the fluctuation-dissipation theorem holds, the spectrum of the collective fluctuations is written

$$ \int _{-\infty }^{+\infty } <v_\mathbf {q}(0) v_\mathbf {q}(t)> e^{-i\omega t} dt= \frac{-2k_BT}{\omega } \chi "(\mathbf {q}, \omega , T).$$

The mean square amplitude of the fluctuations is directly obtained:

$$ <|v_\mathbf {q}^2|>\,= \frac{1}{2\pi }\int \frac{-2k_BT}{\omega } \chi "(\mathbf {q}, \omega , T) d\omega = k_BT \chi (\mathbf {q}, 0, T) $$

At $T= T_c$ the amplitude of the fluctuations for the collective component at $\mathbf {q}_c$ diverges. In other words, on cooling towards $T_c$ larger and larger fluctuations become correlated over longer times until at the instability limit one has long-range fluctuations at zero frequency.¹⁵

Problems

Problem 15.1

In the light of the Clapeyron equation discuss the slope of the function P versus T of the solid -liquid line for water in comparison to the one for all the other systems (see Figs. 15.1 and 15.3), explaining why ice is required in order to allow skating rather than using a floor of any solid material.

Solution: The particular behaviour of ice versus water is related to the fact that ice has a specific volume reduced with respect to water. Thus $V_2< V_1$ (return to Fig. 15.2 caption) and from the Clapeyron equation $[dP/dT]< 0$. The pressure due to the sharp blade of the skater melts the surface of the ice, thus allowing a smooth sliding.

Another aspect related to the particular behaviour of water is the fact that rivers and lakes freeze from the top.

Problem 15.2

Derive an approximate expression of the entropy jump at the boiling process of water by resorting to the definition of entropy in terms of the number of microstates corresponding to a given macrostate (estimate inspired from the book by Blundell and Blundell ).

Solution: The number $\varOmega$ of microstates (states that can be labelled by the wave vector $\mathbf {k}$) can be related to the volume occupied by a given quantity, in analogy to the cases of the normal modes of the radiation (see Problem 1.25) or of the electronic states in crystals (see Sect. 12.5). For a mole of vapour or of liquid, the ratio of the specific number of states can be written

$$\varOmega _{vap}/ \varOmega _{liq}= (V_{vap}/ V_{liq})^{N_A}\sim (1000)^{N_A},$$

having assumed a ratio between vapour and liquid densities around 10$^3$. Then, from $S= k_B ln\varOmega$, the jump of the entropy turns out

$$\varDelta S= \varDelta (k_Bln\varOmega )\sim k_B ln(1000)^{N_A}\sim 7 R. $$

The Trouton rule (see footnote 1) is usually written $L_{vap}\simeq 10 RT_b$. A remarkable violation is found for Helium, where $L_{vap}/RT_b= 2.4$, an indication of quantum effects.

Problem 15.3

From the equation of motion of a single normal mode Q(t) with an effective elastic constant linearly temperature dependent, by assuming a local electric field proportional to the polarization and then to Q, show that instability can occur, with a frequency approaching zero (return to Sect. 10.6). This problem somewhat anticipates some issues to be described at Chap. 16 (see Problem 16.5).

Solution: The equation of motion is

$$ \mu \ddot{Q}(t) + (a + bT) Q(t)= q E_{loc} = \alpha Q(t), $$

(which pertains to a transverse optical mode, see Sect. 16.3). $\mu$ is a reduced mass. Thus

$$ \mu \ddot{Q}(t)+ b[ T- (\alpha - a)/b ]Q(t)= 0 $$

and the effective frequency becomes $\varOmega ^2(T)= ( b/\mu ) [T- T_c]$ with $T_c= (\alpha -a)/b$. In order to have instability $T_c$ must be positive.

In crystals the electric polarization associated with a transverse optical mode at zero wave vector is the sum of an ionic term NqQ (q the ionic charge) plus the term due to the local field $E_{loc}= 4\pi P/3$. In general a small correction to the short range elastic constant occurs (this justifies why at Sect. 14.3.2 it has been neglected). Transition of displacive character to a ferroelectric phase might occur when the proportionality factor between $E_{loc}$ and P is larger than the Lorentz factor $4\pi /3$ (see Sect. 16.2).

Problem 15.4

Starting from the fluctuation-dissipation relationship and by resorting to Kramers-Kronig relationships derive the expression for the paramagnetic static uniform susceptibility in the high temperature limit ($T \gg J$, the exchange interaction).

Solution: The Kramers-Kronig relationships are

$$\begin{aligned} \chi '(\mathbf {q},\omega )-\chi '(\mathbf {q},\infty )= \frac{1}{\pi }\mathcal {P}\int \frac{\chi ''(\mathbf {q}, \omega ')}{\omega '- \omega }d\omega '\ \end{aligned}$$

$$\begin{aligned} \chi ''(\mathbf {q},\omega )=- \frac{1}{\pi }\mathcal {P}\int \frac{\chi ' (q, \omega ')}{\omega '- \omega }d\omega '\ \end{aligned}$$

(for remind see Sect. 16.1).

From the first equation by expressing the imaginary part of the spin susceptibility in terms of the DSF (see Sect. 15.4 and Eq. (15.36)), one notices that for $\omega \rightarrow 0$

$$\begin{aligned} \chi '(0,0)= & {} \frac{1}{\pi }\mathcal {P}\int \frac{\chi ''(0,\omega ')}{\omega '}d\omega ' \\= & {} \frac{1}{\pi }\mathcal {P} \int {d\omega '}{\omega '}S_{\alpha \alpha }(0,\omega ')\frac{e^{\beta \hbar \omega }-1}{2 e^{\beta \hbar \omega }}, \mathrm { with }\,\,\beta = \frac{1}{k_BT} \end{aligned}$$

For $\beta \hbar \omega \ll 1$ one can write

$$\begin{aligned} \chi '(0,0)= & {} \frac{1}{\hbar }\int \frac{1}{k_{B}T}\frac{\hbar \omega d\omega }{\omega }\int dt e^{i\omega t}<S_{0} ^\alpha (t) S_{0} ^{\alpha } (0)>\,= \\= & {} \frac{1}{k_{B}T}\int dt \delta (t=0)<S_{0} ^\alpha (t) S_{0} ^{\alpha } (0)>\,= \\= & {} \frac{1}{k_{B}T}<\left| \sum _{i}S_{i} ^\alpha \right| ^2>\\ \end{aligned}$$

For $k_{B}T \gg J$ the spins are uncorrelated and one has

$$ \chi '(0,0)=\sum _{i}\frac{<\left| S_{i} ^\alpha \right| ^2>}{k_{B}T}= \frac{S(S+1)}{3k_{B}T}, $$

having assumed isotropic spin fluctuations (return to Problem 4.10).

Problem 15.5

Show that the isothermal compressibility of a fluid is proportional to the fluctuation of the mean square number of particles.

Solution: In a grand canonical ensemble the fluctuation in the total number of particles N is

$$\begin{aligned}<(N-<N>)^2>= & {} \frac{1}{Z}\sum ^{\infty }_{N=0} \frac{1}{N! h^{3N}}\int d\mathbf {r}\, d\mathbf {p}\, N^2 e^{-\beta U_{N}} e^{\beta \mu N}-\\- & {} \left[ \frac{1}{Z}\sum ^{\infty }_{N=0} \frac{1}{N! h^{3N}}\int d\mathbf {r}d\mathbf {p}\, N e^{-\beta U_{N}} e^{\beta \mu N}\right] ^2 \\= & {} (k_{B}T)^2\frac{\partial ^2 lnZ}{\partial \mu ^2}. \end{aligned}$$

Since $PV/k_{B}T=lnZ$ (see the book by Amit and Verbin).

$$\begin{aligned}<(N-<N>)^2>\,=(k_{B}T)^2\left[ \frac{\partial ^2 PV/k_{B}T}{\partial \mu ^2}\right] _{T,V}=k_{B}TV \left[ \frac{\partial ^2 P}{\partial \mu ^2}\right] _{T,V} \end{aligned}$$

Now $\left[ \frac{\partial P}{\partial \mu }\right] _{T,V}=\frac{<N>}{V}=n$, so that

$$\begin{aligned}<(N-<N>)^2>= & {} k_{B}TV \left[ \frac{\partial (\frac{<N>}{V})}{\partial \mu }\right] _{T,V}\,=-\frac{<N>k_{B}TV}{V^2}\left[ \frac{\partial V}{\partial \mu }\right] _{T,N}. \end{aligned}$$

By recalling that $K_{T}=-\frac{1}{V}\left[ \frac{\partial V}{\partial \mu }\right] _{T,N}$

$$\begin{aligned}<(N-<N>)^2>\,= \frac{<N>^2k_{B}T}{V}K_{T}=\,<N>nk_{B}TK_{T}. \end{aligned}$$

In ideal gas $PV = Nk_{B}T$ and then

$$\begin{aligned} K^0_{T}=-\frac{1}{V}\frac{\partial (Nk_{B}T/P)}{\partial P}=\frac{1}{V}\frac{1}{P^2}Nk_{B}T=\frac{V}{N}\frac{1}{k_{B}T}=\frac{1}{nk_{B}T} \end{aligned}$$

and finally

$$\begin{aligned} \frac{K_{T}}{K^0_{T}}=\frac{<(N-<N>)^2>}{<N>}. \end{aligned}$$

For detailed description see the books by Stanley and by Amit and Verbin.

Problem 15.6

Derive the relationship between the isothermal compressibility and the density correlation function in fluids. Then, show that the intensity of the radiation scattered with momentum $\mathbf q$ is directly proportional to the static structure factor.

Solution: Since

$$\begin{aligned}<(N-<N>)^2>\,= & {} <\int d \mathbf {r}(n(\mathbf {r})-<n(\mathbf {r})>)\int d\mathbf {r'} \, \left( n(\mathbf {r'})-<n(\mathbf {r'})>\right) > \\= & {} \int \int d \mathbf {r}d \mathbf {r'}g(\mathbf {r}-\mathbf {r'})= V \int g(\mathbf {R}) d\mathbf {R}, \end{aligned}$$

while for a fluid $g(\mathbf {r'}-\mathbf {r})=\,<(n(\mathbf {r'})-<n>)(n(\mathbf {r})\,{-} <n>)>$, according to Problem 15.5

$$\begin{aligned} \frac{K_{T}}{K^0_{T}}=\frac{1}{n}\int g(\mathbf {r}) d\mathbf {r}. \end{aligned}$$

The behaviour of the correlation function can be studied by means of radiation scattering techniques. The intensity of the scattered radiation, after having exchanged a wavevector $\mathbf {q}$ with the fluid, is

$$\begin{aligned} I(\mathbf {q})= \,<\left| \sum _{j}a_{j}(\mathbf {q})\right| ^2> , \end{aligned}$$

where $a_{j}(\mathbf {q})=a e^{-i\mathbf {q}\cdot \mathbf {r_{j}}}$ is the amplitude of the radiation scattered at wave vector $\mathbf {q}$ by the molecule at site j. Then $I(\mathbf {q})=a^2<\left| \sum _{j=1}^N e^{-i\mathbf {q}\cdot \mathbf {r_{j}}}\right| ^2>$, in the case of uncorrelated molecules leading to $I^0(\mathbf {q})=Na^2$. Thus

$$\begin{aligned} \frac{I(\mathbf {q})}{I^0(\mathbf {q})}= & {} \frac{1}{N}<\sum _{i,j}e^{-i\mathbf {q}\cdot (\mathbf {r_{i}}-\mathbf {r_{j}})}> \nonumber \\= & {} \frac{1}{N}\int \int d\mathbf {r} d\mathbf {r'}<\sum _{i,j}\delta (\mathbf {r}-\mathbf {r_{i}})\delta (\mathbf {r'}-\mathbf {r_{j}})e^{-i\mathbf {q}\cdot (\mathbf {r}- \mathbf {r'})}>\nonumber \\= & {} \frac{1}{N}\int \int d\mathbf {r} d\mathbf {r'}e^{-i\mathbf {q}\cdot (\mathbf {r}-\mathbf {r'})}<\underbrace{\sum _{i}\delta (\mathbf {r}-\mathbf {r_{i}})}_{n(r)}\underbrace{\sum _{j}\delta (\mathbf {r'}-\mathbf {r_{j}})}_{n(r')}> \nonumber \\= & {} \frac{1}{N}\int \int d\mathbf {r} d\mathbf {r'}e^{-i\mathbf {q}\cdot (\mathbf {r}-\mathbf {r'})}\left[ g(\mathbf {r}-\mathbf {r'})+n^2\right] . \end{aligned}$$

Namely

$$\begin{aligned} \frac{I(\mathbf {q})}{I^0(\mathbf {q})}=\frac{1}{n} \int d\mathbf {R}g(\mathbf {R}) e^{i\mathbf {q}\cdot \mathbf {R}} + \frac{V^2}{N}n^2\delta (\mathbf {q}). \end{aligned}$$

The first term, the Fourier transform of the correlation function at wavevector $\mathbf {q}$, is the static structure factor $S(\mathbf {q})$ which gives the amplitude of the collective modes at wave-vector $\mathbf {q}$

$$\begin{aligned} S(\mathbf {q})= \int d\mathbf {R}g(\mathbf {R}) e^{i\mathbf {q}\cdot \mathbf {R}}\propto <|m_\mathbf {q}|^2> \propto (1+ q^2\xi ^2)^{-1} \,\,\, \end{aligned}$$

(see Eq. (15.40)). The compressibility, being the response function to a uniform $q=0$ perturbation, is directly proportional to $S(q=0)$.

As illustrative example, the neutron scattering intensity is reported above as a function of the wavevector q exchanged with D$_2$O on approaching the critical temperature $T_c\simeq 637$ K (left). The enhancement of the fluctuations at $q=0$ is detected. The inverse intensity is reported as a function of $q^2$ in the plot on the right and a linear trend is observed, as expected. The intercept is proportional to the inverse correlation length. The decrease in the intercept on approaching $T_c$ evidences the divergence in the correlation length.

Problem 15.7

From the equation of motion for underdamped oscillator, derive the dynamical structure factor.

Solution: For underdamped normal oscillators, i.e. such that $\varOmega _\mathbf {q}^2\ll 4 k_\mathbf {q} M$, the solution of Eq. (15.33) is

$$\begin{aligned} m_\mathbf {q}(t)= e^{-\frac{\varOmega _\mathbf {q}t}{2M}}\biggl ( m_\mathbf {q}(0)cos\omega _\mathbf {q}t + \biggl [ \frac{\partial m_\mathbf {q}(t)}{\partial t} + \frac{m_\mathbf {q}(0)\varOmega _\mathbf {q}}{2M}\biggr ] \frac{sin\omega _\mathbf {q}t}{\omega _\mathbf {q}}\biggr ), \end{aligned}$$

with

$$\begin{aligned} \omega _\mathbf {q}= \biggl ( \frac{k_\mathbf {q}}{M} - \frac{\varOmega _\mathbf {q}^2}{4M^2}\biggr )^{1/2}. \end{aligned}$$

yielding for the correlation function

$$\begin{aligned} g_\mathbf {q}(t)= <|m_\mathbf {q}(0)|^2> e^{-{\varOmega _\mathbf {q}t}/{2M}}\biggl ( cos\omega _\mathbf {q}t + \biggl [ \frac{\varOmega _\mathbf {q}}{2M\omega _\mathbf {q}}\biggr ] sin\omega _\mathbf {q}t \biggr ). \end{aligned}$$

Due to the deterministic character of the motion the statistical ensemble average for the correlation function only involves the initial condition, i.e. the value of $m_\mathbf {q}(0)$. In the equation above one can set $\partial m_\mathbf {q}(0)/\partial t= 0$ since the recovery of the collective order parameter towards equilibrium, after a fluctuation, initiates with zero velocity.

The mean square value is related to the temperature by the average over the initial amplitude $A_\mathbf {q}$:

$$\begin{aligned} <|m_\mathbf {q}(0)|^2>= \frac{1}{2A_\mathbf {q}}\int _{-A_\mathbf {q}}^{+A_\mathbf {q}} m_\mathbf {q}(0)^2 dm_\mathbf {q}= \frac{A_\mathbf {q}^2}{3}= \frac{2k_BT}{3k_\mathbf {q}}. \end{aligned}$$

Thus from the Fourier transform of $g_\mathbf {q}(t)$ one obtains

$$\begin{aligned} S(\mathbf {q},\omega )= \frac{2k_BT}{3k_\mathbf {q}} \frac{{4\varOmega _\mathbf {q}\omega _\mathbf {q}^2}/{M}}{(\omega _\mathbf {q}^2-\omega ^2)^2 + (4 \varOmega _\mathbf {q}^2\omega _\mathbf {q}^2/M)}. \end{aligned}$$

corresponding to Eq. (15.34).